8. String to Integer (atoi)

LeetCode
,

My Solution (Flags)

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class Solution {
    public int myAtoi(String s) {
        int length = s.length();
        int sign = 1;
        long result = 0;
        boolean acceptSpace = true;
        boolean acceptSign = true;
        boolean acceptZero = true;
        for (int i = 0; i < length; i++) {
            // get char
            char currentChar = s.charAt(i);
            // process whitespace
            if (currentChar == ' ' && acceptSpace) {
                continue;
            } else if (currentChar == ' ' && !acceptSpace) { 
                break;
            }
            // process sign
            else if (currentChar == '+' && acceptSign) {
                acceptSpace = false;
                acceptSign = false;
                continue;
            }

            else if (currentChar == '-' && acceptSign) {
                acceptSpace = false;
                acceptSign = false;
                sign = -1;
                continue;
            }

            else if (currentChar == '0' && acceptZero) {
                acceptSpace = false;
                acceptSign = false;
                continue;
            }

            else if (Character.isDigit(currentChar)) {
                acceptSpace = false;
                acceptSign = false;
                acceptZero = false;
                
                int digit = currentChar - '0';
                if (result > Integer.MAX_VALUE / 10 || 
                   (result == Integer.MAX_VALUE / 10 && digit > (sign == 1 ? 7 : 8))) {
                    return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
                }

                result = result * 10 + digit;
            }

            else {
                break;
            }
        }

        result = result * sign;
        if (result > Integer.MAX_VALUE) return Integer.MAX_VALUE;
        if (result < Integer.MIN_VALUE) return Integer.MIN_VALUE;
        return (int) result;

    }
}

Remarks:

  1. Check the last digit of MAX value of int boundary: result == Integer.MAX_VALUE / 10 && digit > (sign == 1 ? 7 : 8))

  2. Use long to store the intermediate result, and check if the result is out of bound at last:

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    if (result > Integer.MAX_VALUE) return Integer.MAX_VALUE;
    if (result < Integer.MIN_VALUE) return Integer.MIN_VALUE;
    return (int) result;

While Loops

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public class Solution {
    public int myAtoi(String s) {
        int i = 0, n = s.length();
        int sign = 1;
        int result = 0;

        // Step 1: Skip leading whitespaces
        while (i < n && s.charAt(i) == ' ') {
            i++;
        }

        // Step 2: Check sign
        if (i < n && (s.charAt(i) == '+' || s.charAt(i) == '-')) {
            sign = s.charAt(i) == '-' ? -1 : 1;
            i++;
        }

        // Step 3: Read digits
        while (i < n && Character.isDigit(s.charAt(i))) {
            int digit = s.charAt(i) - '0';

            // Step 4: Handle overflow
            if (result > (Integer.MAX_VALUE - digit) / 10) {
                return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }

            result = result * 10 + digit;
            i++;
        }

        return result * sign;
    }
}

Remarks:

  1. result * 10 + digit > Integer.MAX_VALUE -> result > (Integer.MAX_VALUE - digit) / 10 Use this to check out of bound